Find the Limit.
lim x->oo sqrt(x^2+1)
oo=infinity
It does not. It use "$\displaystyle \geq$" there.
Basically I am showing that $\displaystyle \sqrt{x^2+1}$ is larger than $\displaystyle x$. And since $\displaystyle x$ goes to $\displaystyle \infty$ must mean $\displaystyle \sqrt{x^2+1}$ goes to $\displaystyle \infty$.